Question: Graph this system of equations and solve. $12x+6y = -30$ $y = \dfrac{2}{3} x + 3$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Answer: Convert the first equation, $12x+6y = -30$ , to slope-intercept form. $y = -2 x - 5$ The y-intercept for the first equation is $-5$ , so the first line must pass through the point $(0, -5)$ The slope for the first equation is $-2$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move down (because it's negative) $1$ position to the right. $2$ positions down from $(0, -5)$ is $(1, -7)$ Graph the blue line so it passes through $(0, -5)$ and $(1, -7)$ The y-intercept for the second equation is $3$ , so the second line must pass through the point $(0, 3)$ The slope for the second equation is $\dfrac{2}{3}$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move up $3$ positions to the right. $2$ positions up from $(0, 3)$ is $(3, 5)$ Graph the green line so it passes through $(0, 3)$ and $(3, 5)$ The solution is the point where the two lines intersect. The lines intersect at $(-3, 1)$.